VIPSolutionsA 2 8 5 g silver figure o f a polar bear i s dropped into the 2 2 1 g aluminum cup o f a well - insulated calorimeter containing 2 6 9 g o f liquid water a t 2 4 . 5 ° C . The bear's initial temperature i s 9 7 . 9 ° C . The specific heats o f silver, aluminum, and liquid water are, respectively, 2 3 4 J k g * K , 9 1 0 J k g * K , and 4 1 9 0 J k g * K . What i s the final temperature T e q o f the water, cup, and bear when they reach thermal equilibrium? T e q =
Question:
A 2 8 5 g silver figure o f a polar bear i s dropped into the 2 2 1 g aluminum cup o f a well - insulated calorimeter containing 2 6 9 g o f liquid water a t 2 4 . 5 ° C . The bear's initial temperature i s 9 7 . 9 ° C . The specific heats o f silver, aluminum, and liquid water are, respectively, 2 3 4 J k g * K , 9 1 0 J k g * K , and 4 1 9 0 J k g * K . What i s the final temperature T e q o f the water, cup, and bear when they reach thermal equilibrium? T e q =
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Asked by: Test User 100093
Created at: 2025-04-28 03:53:00
Asked by: Test User 100093
Created at: 2025-04-28 03:53:00
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Let's solve the calorimetry problem step by step.
### **Given:**
- **Silver bear:**
Mass (\( m_{Ag} \)) = 285 g = **0.285 kg**
Initial temperature (\( T_{i,Ag} \)) = **97.9°C**
- **Aluminum cup:**
Mass (\( m_{Al} \)) = 221 g = **0.221 kg**
Initial temperature (\( T_{i,Al} \)) = **24.5°C**
- **Water:**
Mass (\( m_{H_2O} \)) = 269 g = **0.269 kg**
Initial temperature (\( T_{i,H_2O} \)) = **24.5°C**
- **Specific heats:**
\( c_{Ag} = 234 \) J/(kg·K)
\( c_{Al} = 910 \) J/(kg·K)
\( c_{H_2O} = 4190 \) J/(kg·K)
---
### **Assumptions**
- No heat loss to environment (well-insulated calorimeter).
- All components reach **final equilibrium temperature \( T_{eq} \)**.
---
### **Set Up the Equation**
Heat lost by hot object (silver bear) = Heat gained by cold objects (aluminum cup + water):
\[
Q_{lost} + Q_{gained} = 0
\]
\[
(m_{Ag} c_{Ag}) (T_{eq} - T_{i,Ag}) + (m_{Al} c_{Al}) (T_{eq} - T_{i,Al}) + (m_{H_2O} c_{H_2O}) (T_{eq} - T_{i,H_2O}) = 0
\]
Note: \( (T_{eq} - T_{i,Ag}) \) will be negative since the bear cools down, and the other two will be positive.
---
\[
(0.285)(234)(T_{eq} - 97.9) + (0.221)(910)(T_{eq} - 24.5) + (0.269)(4190)(T_{eq} - 24.5) = 0
\]
---
### **Calculate the coefficients:**
#### **Silver bear:**
\[
0.285 \times 234 = 66.69
\]
#### **Aluminum cup:**
\[
0.221 \times 910 = 201.11
\]
#### **Water:**
\[
0.269 \times 4190 = 1127.11
\]
---
### **Expand the equation:**
\[
66.69(T_{eq} - 97.9) + 201.11(T_{eq} - 24.5) + 1127.11(T_{eq} - 24.5) = 0
\]
---
\[
66.69 T_{eq} - 6530.451 + 201.11 T_{eq} - 4927.195 + 1127.11 T_{eq} - 27613.195 = 0
\]
---
\[
(66.69 + 201.11 + 1127.11) T_{eq} - (6530.451 + 4927.195 + 27613.195) = 0
\]
\[
1394.91 T_{eq} - 39070.841 = 0
\]
---
### **Solve for \( T_{eq} \):**
\[
1394.91 T_{eq} = 39070.841
\]
\[
T_{eq} = \frac{39070.841}{1394.91}
\]
\[
T_{eq} \approx 28.0^\circ C
\]
---
## **Final Answer:**
\[
\boxed{T_{eq} \approx 28.0^\circ \text{C}}
\]
---
**Summary of steps:**
1. Set up conservation of energy.
2. Plug in masses, specific heats, and temperatures.
3. Solve for the final equilibrium temperature.
Let me know if you want the answer with more significant digits or if you want to see the calculation in Kelvin (though temperature changes in °C and K are the same).
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